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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should lift be accelerated upwards in order to reduce its period to T/2? (g is acceleration due to gravity).

    A)  2 g                                        

    B)  3 g

    C)  4 g                                        

    D)  g

    Correct Answer: B

    Solution :

    Time period of simple pendulum is given by                 \[T=2\pi \sqrt{\frac{l}{g}}\]                                          ... (i) When the lift is moving up with an acceleration a, then time period becomes                 \[T'=2\pi \sqrt{\frac{l}{g+a}}\] Here,     \[T'=\frac{T}{2}\] \[\Rightarrow \]               \[\frac{T}{2}=2\pi \sqrt{\frac{l}{g+a}}\]                                  ?. (ii) Dividing Eq. (ii) by Eq. (i), we get                 \[a=3g\]


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